Set Up an Integral Representing the Volume of the Torus $(x-6)^{2}+y^{2}=1$

question 12

Multiple Choice

Set up an integral representing the volume of the torus (donut) obtained by rotating the circle $(x-6) ^{2}+y^{2}=1$ about the y-axis.

• A) \(\int_{-1}^{1} \pi\left(6+\sqrt{1-y^{2}}\right)^{2} d y\)
• B) \(\int_{-1}^{1} \pi\left(6-\sqrt{1-y^{2}}\right)^{2} d y\)
• C) \(\int_{-1}^{1} \pi\left(\left(6+\sqrt{1-y^{2}}\right)^{2}+\left(6-\sqrt{1-y^{2}}\right)^{2}\right) d y\)
• D) \(\int_{-1}^{1} \pi\left(\left(6+\sqrt{1-y^{2}}\right)^{2}-\left(6-\sqrt{1-y^{2}}\right)^{2}\right) d y\)

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